Official DAT Destroyer Q&A Thread

[densaugeo]

Hi guys, since I'm currently going through the DAT Destroyer and I'm sure many of you guys are as well I felt this thread would be helpful. I usually have various questions while going through Destroyer regarding why a certain answer choice is correct or other times I am in need of a more detailed explanation.

Instead of making multiple threads each time we have questions, I thought we could just post them here and anyone can chime in with an explanation or further clarification.

Please don't post entire questions as that would be a violation of copyright. You can ask specific questions regarding a particular problem in the Destroyers.

Hopefully others find this a useful thread. Thanks!

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[DAT Destroyer]

2017 DAT Destroyer
Biology Question #297

Is the answer 4N during anaphase because a somatic cell is already 2N (46 chromosomes) so when the sister chromosomes split and the chromosome # is now 96 you do 2N x 2? I answered 2N because the chromosome number of somatic cells is 46 and during anaphase it is 96. I am assuming I was supposed to consider that they were already 2N but I would like to make sure.

Yes, that is correct. Your stated reasoning is also correct.

 

[DAT Destroyer]

Hello Dr Romano,

I have a question on the DAT Destroyer 2017 edition, General Chemistry #50.

Although I agree, that both acids and their corresponding salt are in a 1:1 ratio, thus according to the Henderson Hasselbach, pH = pKa + Log (1) --- which is 0... Thus the equation becomes pH = pKa. However, without having the Ka or pKa values of the acids, how can we conclude that their pH values will be the same, if pH = pKa?

The is a MAJOR concept. If the acid and salt are in a 1 to 1 ratio,,,,this the HALF EQUIVALENCE point.....at this point, we have the GREATEST BUFFER.......thus pH = pKa. This fact will be vital when you go to dental school and sitting in BioChemistry class !!!!!
If you simply write out the Henderson Hasselbach equation.....we see the log term cancel.....hence you are left with the expression pH = pKa. I hope this helps. For further clarity, the text by Raymond Chang is my favorite.

Hope this helps.


Dr. Jim Romano

 

[spdental]

Dr. Romano,

On question 75 of 2017 Ochem Section, I was wondering why HCl would shift? Isn't this a primary alcohol (wouldn't it preform an SN2 mech)?

 

[ColoradoTeeth]

Hey Dr. Romano, for Orgo #204 (2016 DAT Destroyer), on the actual dat, would they give us the pka values of an amino acid or are we suppose to know it?

you definitely don't have to know AA pka values

 

[deleted768895]

The is a MAJOR concept. If the acid and salt are in a 1 to 1 ratio,,,,this the HALF EQUIVALENCE point.....at this point, we have the GREATEST BUFFER.......thus pH = pKa. This fact will be vital when you go to dental school and sitting in BioChemistry class !!!!!
If you simply write out the Henderson Hasselbach equation.....we see the log term cancel.....hence you are left with the expression pH = pKa. I hope this helps. For further clarity, the text by Raymond Chang is my favorite.

Hope this helps.


Dr. Jim Romano




.

Thank you for your response. I understand that the acid and conjugate base are in a 1:1 ratio and that pH = pKA, however, the answer says that Buffer A and Buffer B have the same pH, and I dont know how that conclusion has been made. Buffer A consists of 2M HOAc and 2M NaOAc (thus 1:1, pH = pKa)) and Buffer B consists of 0.04M HOAc and 0.04M NaOAc (again 1:1, its pH = pKa). My question is, how can we conclude that the pH of both buffers is the same, if we do not have the pKa of either acids? if pH = pKa surely we need pKa to determine this???

 

[DAT Destroyer]

Thank you for your response. I understand that the acid and conjugate base are in a 1:1 ratio and that pH = pKA, however, the answer says that Buffer A and Buffer B have the same pH, and I dont know how that conclusion has been made. Buffer A consists of 2M HOAc and 2M NaOAc (thus 1:1, pH = pKa)) and Buffer B consists of 0.04M HOAc and 0.04M NaOAc (again 1:1, its pH = pKa). My question is, how can we conclude that the pH of both buffers is the same, if we do not have the pKa of either acids? if pH = pKa surely we need pKa to determine this???.

The conclusion was made as follows,,,,Diluting the buffer will NOT change the pH because the RATIO stays the same. Say we have 1M and 1M of acid and its salt....we dilute it by a factor of 10.......we now have 0.1 and 0.1,,,,,thus when you put this into the Henderson Hasselbach equation, we still take the log of 1 which is zero. We then end it at the half equivalence point at which pH is equal to pKa. With that said, dilution will however DECREASE the buffer capacity. I hope this helps.

 

[deleted768895]

Thank you for your response. I understand that the acid and conjugate base are in a 1:1 ratio and that pH = pKA, however, the answer says that Buffer A and Buffer B have the same pH, and I dont know how that conclusion has been made. Buffer A consists of 2M HOAc and 2M NaOAc (thus 1:1, pH = pKa)) and Buffer B consists of 0.04M HOAc and 0.04M NaOAc (again 1:1, its pH = pKa). My question is, how can we conclude that the pH of both buffers is the same, if we do not have the pKa of either acids? if pH = pKa surely we need pKa to determine this???.

OH wait a minute, is Buffer A and Buffer B made from the same acid/salt?!

 

[deleted768895]

.
The conclusion was made as follows,,,,Diluting the buffer will NOT change the pH because the RATIO stays the same. Say we have 1M and 1M of acid and its salt....we dilute it by a factor of 10.......we now have 0.1 and 0.1,,,,,thus when you put this into the Henderson Hasselbach equation, we still take the log of 1 which is zero. We then end it at the half equivalence point at which pH is equal to pKa. With that said, dilution will however DECREASE the buffer capacity. I hope this helps.

WOW I just realised my confusion was from the fact that I thought Buffer A and Buffer B are completely different acids/conjugate base!

 

[yoh95]

Dr. Romano OC Odyssey chapter 7 #36 I believe carbon dioxide should have a linear geometry

 

[Dallas Dentist]

2017 DAT Destroyer QR Section:
Question #146
I am not understanding how the solution is explaining the answer to this question. Can anyone please explain?

 

[DAT Destroyer]

Dr. Romano OC Odyssey chapter 7 #36 I believe carbon dioxide should have a linear geometry

Yes......C02 is linear. The question asks for a TRIPLE bond,,,,,This is seen in CO and C2H2........Hope this helps.

 

[yoh95]

For Grignard reagents: Why do acidic protons have preference over the ketone if they are present in the same compound?

 

[DAT Destroyer]

For Grignard reagents: Why do acidic protons have preference over the ketone if they are present in the same compound?

You have asked a great question that is often debated in many Advanced Organic Chemistry lectures. Let us look at this from the most simplest perspective. As a good rule of thumb, the ACID -BASE reaction occurs faster than the carbonyl attack. If an acidic proton is present such as a carboxy acid or alcohol, the Grignard will preferentially remove this proton. By doing this, the carbonyl bond is maintained and is not destroyed. The interaction with the carbon nucleophile ( Grignard ) with a proton's antibonding orbital is lower in energy than the direct attack on the antibonding carbonyl orbital of the aldehyde or ketone. The reaction runs along the easier path, and that means abstraction of a hydrogen. Thus, compounds like alcohols, acids,alkynes and thiols easily react with a Grignard. In practice, this is often a big problem. We have what is called protecting groups that we employ to mask these functionalities. There are books on protecting groups, that you don't want to see. The last one that I read was nearly 1000 pages long. Not a week goes by that another new and improved protecting group comes along or one becomes obsolete. The bottom line to all this madness is as follows,,,,,,The acid base reaction occurs faster if the carbonyl group can be maintained.

I hope this helps.

Dr. Romano

 

[spdental]

For question 119 on Ochem 2017, I was wondering when you use the N- in nomenclature? I can't find anything online.

Also for question 75, it claims that there would be a shift with HCl, however once the OH is protanted, wouldn't this go through an SN2 mech and not form a carbocation? It is a primary alcohol group (or is H2O a good enough leaving group to force this into SN1?)

Thanks.

 

[mrdeez]

For question 119 on Ochem 2017, I was wondering when you use the N- in nomenclature? I can't find anything online.

Also for question 75, it claims that there would be a shift with HCl, however once the OH is protanted, wouldn't this go through an SN2 mech and not form a carbocation? It is a primary alcohol group (or is H2O a good enough leaving group to force this into SN1?)

Thanks.

For question 75, his argument for that happening is a "concerted alkyl shift" that was alluded to in one of the Klein orgo books. The Klein book briefly mentions the orbital overlap in neopentanol and how this overlap will lead to a reaction that is not SN2. However, Dr. Romano is wrong in using this example to explain his reasoning, because the molecule in the destroyer example does not experience the same orbital overlap. If a concerted mechanism were to occur like he says it would, it would be a concerted hydride shift, which is not documented in any orgo book as being energetically favorable.

SOCl2 and HCl would then be equally useful options on the basis of their mechanism. But the question asks for THE BEST method. If two reagents create the same product with equal efficiency, organic synthesis laboratories tend to use the cheaper, more readily available option. In this problem, SOCl2 is more expensive than HCl when purchasing for a lab. So not only is he wrong in his argument regarding the mechanism, the SOCl2 is still not the best option due to the availability and low cost of HCl.

His reasoning for the entire problem is SO nuanced that I don't think its anything you need to spend much time on for the DAT, imo.

TLDR; you are completely right, his explanation is debatable at best. Continue as you were. Godspeed.

Edit: I just realized you don't have my same edition of destroyer. I was referencing #70 in the 2016 edition and I assumed yours was just a different number. Mine asks: Which reagents would be best for this reaction: (picture of 2-methylpropanol goes to 1-chloro-2-methylpropanol) a. Cl2, hu b. SOCl2, pyridine c. HCl d. HOCl e. B and C
Is yours the same?

 

[TigerDK]

2017 Destroyer OChem #7

This may seem like a stupid question, but why does OH give up its H more easily than NH2?

 

[spdental]

2017 Destroyer OChem #7

This may seem like a stupid question, but why does OH give up its H more easily than NH2?

Greater electronegativity on the oxygen forces the electrons to pull in closer to the oxygen's nucleus, effectively leaving a hydrogen that has "partially positive" character and will give up H easier. Large atoms do the same but they can spread the charge out even better due to their large size.

 

[DAT Destroyer]

2017 Destroyer OChem #7

This may seem like a stupid question, but why does OH give up its H more easily than NH2?

The Pka of any alcohol is much much lower than that of an amine. Phenol has a pKa of about 10, an aromatic amine more than double this. The lower the pKa the more willing will it be to donate a proton, and act as an acid. Thus , this means that a base would more readily form ! As a good general rule, keep saying to yourself what I tell my students....AMINES are usually BASES...and do not want to lose an H. An NH2 group for your purposes normally does NOT lose a hydrogen. !!!!!

Hope this helps.

Dr. Romano

 

[dl2501]

The concept is this......the more particles of solute that is dissolved in the solvent will cause the freezing point to lower, and the boiling point to be raised. Calcium acetate gives a total of 3 particles.....one Ca++ and 2 Acetate ions for a total of 3 particles . Sodium acetate only gives 2 particles...Na+ and acetate. Thus sodium acetate will lower the freezing point but NOT AS MUCH since there are fewer particles dissociated. So we are looking for a value not quite as low as -2.8 degrees. Hope this helps.

Dr. Romano,

Regarding this question, wouldn't it rely also on the Kf because Kf is the temperature change per one molal? I know it has to be a small change, but wouldn't the small change be relative to what the Kf value is?

Thank you in advance.

 

[TigerDK]

The Pka of any alcohol is much much lower than that of an amine. Phenol has a pKa of about 10, an aromatic amine more than double this. The lower the pKa the more willing will it be to donate a proton, and act as an acid. Thus , this means that a base would more readily form ! As a good general rule, keep saying to yourself what I tell my students....AMINES are usually BASES...and do not want to lose an H. An NH2 group for your purposes normally does NOT lose a hydrogen. !!!!!

Hope this helps.

Dr. Romano

Thank you! Another question though haha.
OChem #71 2017 Destroyer

What kind of a reaction is enamine + C2H5I/H2O? I've never seen a reaction like this before.

 

[RDHtoDDS_40]

From 2017 DAT Destroyer General Chemistry section:
Q- #106.) What is the pH of 1.0x10^-4 M Ca(OH)2 solution?
A- 10.3

From answer key:
[OH]=2 (1x10^-4)
pOH=-log [2x10^-4]
pOH= 3.7 QUESTION --------> How is this value of 3.7 derived from -log [2x10^-4]??
Final answer: 14-3.7 = 10.3

@orgoman22

 

[dl2501]

From 2017 DAT Destroyer General Chemistry section:
Q- #106.) What is the pH of 1.0x10^-4 M Ca(OH)2 solution?
A- 10.3

From answer key:
[OH]=2 (1x10^-4)
pOH=-log [2x10^-4]
pOH= 3.7 QUESTION --------> How is this value of 3.7 derived from -log [2x10^-4]??
Final answer: 14-3.7 = 10.3

@orgoman22

For this question, you have to know that 1) Ca(OH)2 is a strong base, 2) 2OH is what dissociates (and you know it has to dissociate completely (because of 1), and 3) (and this is recommended) that you know the 3.16x10^-(whatever) magic number).

I am assuming you know the -log and how that all works out:
1x10^-3 = pH3
1x10^-4 = pH4
1x10^-5 = pH5


We know that because it fully dissociates, the concentration has to x2, and that is where the 2x10^-4 comes in. Now here is the "magic part." Log is not a linear scale, so it can't be 1:1 (just look at the graph). But it comes out that 3.16x10^-(whatever) is equal to " whatever.5 ". How I like to think of it is that if the 1st number of the concentration, in this case 2 (from 2x10^-4) is between 0 and 3.16, then the .X has to be greater than .5" in this case, it is .7 (hence 3.7). Using this same logic, any number that is from 3.16 to 10 will be lower than .5 (i.e. if it was 8x10^-X, i would guesstimate about a X.2 or something.

It is pretty to explain, let me know if it helps!

 

[stu-dent1102]

The answer choices given represent the length and the width. Try each one to see which one will give: L*W=7 and 2(L+W)=12
You don't have to use the property given, you can simply foil and you get the same answer. It's faster if you use the property

Hope this helps, if it did please like my post.

Thanks..

Thank you!
I was not sure if I was supposed to plug in and see what worked or not. I appreciate your help!

 

[spdental]

Dr. Romano,

For question 131 on the Ochem section of 2017, I was wondering why the OH- wouldn't attack the ketone as well? (shouldn't it attack and form a protanted alcohol with the given H+)?

 

[DAT Destroyer]

Dr. Romano,

For question 131 on the Ochem section of 2017, I was wondering why the OH- wouldn't attack the ketone as well? (shouldn't it attack and form a protanted alcohol with the given H+)?

The first move is a slam dunk that will set the stage for the entire sequence, The double alpha has a pKa value of 9, and is the most acidic proton. Attack of the ketone although possible, as you point out competes with the VERY FAST acid base reaction. The acid base reaction almost always wins. This reaction is the famous Acetoacetic ester synthesis....and along with the Malonic ester synthesis represent very important reactions to know. Consult any text for more examples. Hope this helps.

 

[DAT Destroyer]

Thank you! Another question though haha.
OChem #71 2017 Destroyer

What kind of a reaction is enamine + C2H5I/H2O? I've never seen a reaction like this before.

The enamine is simply acting as a NUCLEOPHILE. It is nothing more than an SN2 process. Therefore, when alkylating we usually like to use methyl, primary allylic or benzylic halides. Hydrolysis of this reaction gives the carbonyl compound. Consult any organic text to see a few more examples. If you can look at the mechanism, it would give you a deeper insight to what has occurred, Hope this helps.

 

[spdental]

The first move is a slam dunk that will set the stage for the entire sequence, The double alpha has a pKa value of 9, and is the most acidic proton. Attack of the ketone although possible, as you point out competes with the VERY FAST acid base reaction. The acid base reaction almost always wins. This reaction is the famous Acetoacetic ester synthesis....and along with the Malonic ester synthesis represent very important reactions to know. Consult any text for more examples. Hope this helps.

Thank you.

One more general questions regarding the genetics on the exam:

I'm having difficulty knowing for crossovers when the male uses the Y chromosome in the cross, or if both are X? For example 245 vs 461. Is there a way to differentiate based on what the question is asking?

 

[goprodent2]

Hello Dr. Romano,

Ochem DAT destroyer #68, for figuring out the # of isomers the answer says meso for two of the structures. Does that mean that we would only count the having 2 methyls as dashed or both wedged structures are 1 isomer instead of 2 because they are meso, and we could flip them? or why wouldn't we count the dashed structure for the 2 structures that has meso under them (in answer key)?

Then that brings the question of why wouldn't we just count the 2 middle structures (in the answer key) where one Ch3 is on dash and the other is on a wedge as 1 isomer instead of 2 different ones?

Lastly, when can we use the 2^n rule for stereoisomers? My thought process was there would be 3 total stereocenters so 2^3=8. Then add 1 for having the two methyls on one carbon brings the total isomers to 9. BUT, since there are two meso compounds there would be 7, (I came to this after looking at the answer but I just need help with clarifying how to count meso compounds)

Thank you for your help.

 

[goprodent2]

The enamine is simply acting as a NUCLEOPHILE. It is nothing more than an SN2 process. Therefore, when alkylating we usually like to use methyl, primary allylic or benzylic halides. Hydrolysis of this reaction gives the carbonyl compound. Consult any organic text to see a few more examples. If you can look at the mechanism, it would give you a deeper insight to what has occurred, Hope this helps.

Just a follow-up question, if there wasn't H2O after adding C2H5I, would the answer have been enamine with a double bond shifted to the left, and C2H5 would have added on ortho to the right?

But since water is included, it attacked the double bond first and OH is added. What forces the enamine to leave giving us a carbonyl and C2H5 added?

 

[goprodent2]

Question #77,Ochem 2017 DAT destroyer, it says to do the Aldol condensation, when in dilute acid or base. Do we do this rxn, bc in dilute acid or base, or because we have an alpha carbon?

Why didn't the rxn go through dehydration and create an alkene?

 

[dl2501]

Question #77,Ochem 2017 DAT destroyer, it says to do the Aldol condensation, when in dilute acid or base. Do we do this rxn, bc in dilute acid or base, or because we have an alpha carbon?

Why didn't the rxn go through dehydration and create an alkene?

I had the same question regarding this, and I think it would IF there was heat applied, because the heat is what drives the OH off (I am pretty sure about this). Besides, the answer choices didn't have the alkene in it, which is why I am not going nuts over it.

 

[deleted768895]

From 2017 DAT Destroyer General Chemistry section:
Q- #106.) What is the pH of 1.0x10^-4 M Ca(OH)2 solution?
A- 10.3

From answer key:
[OH]=2 (1x10^-4)
pOH=-log [2x10^-4]
pOH= 3.7 QUESTION --------> How is this value of 3.7 derived from -log [2x10^-4]??
Final answer: 14-3.7 = 10.3

@orgoman22

Hey, I can attempt to answer this question for you....

Ok so say it was just 1x10^-4 then taking -log (1x10^-4) would give a pOH = 4. However 2, is a greater number than 1 (obviously) and therefore 2x10^-4 means there are more OH- ions in solution than there would be if there was 1x10^-4. If there are more OH- ions in solution, the resulting pOH of the solution should therefore be LOWER than 4, and so without a calculator I would've estimated that its around ~3.7-3.8. I guess you have to look at the pH value which is closest to 14-pOH.
Hope this helps!

 

[RDHtoDDS_40]

Hey, I can attempt to answer this question for you....

Ok so say it was just 1x10^-4 then taking -log (1x10^-4) would give a pOH = 4. However 2, is a greater number than 1 (obviously) and therefore 2x10^-4 means there are more OH- ions in solution than there would be if there was 1x10^-4. If there are more OH- ions in solution, the resulting pOH of the solution should therefore be LOWER than 4, and so without a calculator I would've estimated that its around ~3.7-3.8. I guess you have to look at the pH value which is closest to 14-pOH.
Hope this helps!

Yes totally this makes sense. I forgot about this estimation that we can do. Thank you!!

 

[RDHtoDDS_40]

For this question, you have to know that 1) Ca(OH)2 is a strong base, 2) 2OH is what dissociates (and you know it has to dissociate completely (because of 1), and 3) (and this is recommended) that you know the 3.16x10^-(whatever) magic number).

I am assuming you know the -log and how that all works out:
1x10^-3 = pH3
1x10^-4 = pH4
1x10^-5 = pH5


We know that because it fully dissociates, the concentration has to x2, and that is where the 2x10^-4 comes in. Now here is the "magic part." Log is not a linear scale, so it can't be 1:1 (just look at the graph). But it comes out that 3.16x10^-(whatever) is equal to " whatever.5 ". How I like to think of it is that if the 1st number of the concentration, in this case 2 (from 2x10^-4) is between 0 and 3.16, then the .X has to be greater than .5" in this case, it is .7 (hence 3.7). Using this same logic, any number that is from 3.16 to 10 will be lower than .5 (i.e. if it was 8x10^-X, i would guesstimate about a X.2 or something.

It is pretty to explain, let me know if it helps!

This was definitely helpful thank you!! I forgot that we can do this type of estimation

 

[RDHtoDDS_40]

For #138 of the 2017 DAT Destroyer: General Chemistry section:
The Ksp of Fe(OH)2 at 25 degrees C is 1.6x10^-14. What is the solubility of Fe(OH)2 in 0.025 M FeCl2?
From the answer key:
1.6x10^-14=[.025][4x^2]
1.6x10^-14=[2.5x10^-2][4x^2]
1.6x10^-14=[10x10^-2][x^2] ----> QUESTION: from the previous step, did you multiply the numbers by 4 to get these values? Wouldn't you have to manipulate the other side of the equation as well then?
1.6x10^-13=[x^2]
x= 4x10^-7 = final answer

@orgoman22

 

[Dallas Dentist]

#186 OCHEM 2017:

I am not understanding how the monochlorination of 2-methylbutane would give 4 products (not counting stereoisomers). Why wouldn't the methyl group be able to be chlorinated? My guess is that by chlorinating this methyl group, it would yield the same exact product as chlorinating carbon 1, but not sure if that is correct

 

[goprodent2]

2017 DAT Destroyer QR section

(more of a comment) #138 : in the answer key, is it a type where it breaks y^3 into y^2 y^3 in the first radical when it’s simplified? Because in the next two radicals it shows y^2 and y^1.

#146:. I’m just not getting the answer key. I don’t know how in the first step it was reduced to sin(x)/cos (x) which then equals cos (x)/sin(x). From there it says to multiply by the LCD, but multiply to which equation? I’m just lost on this problem

#149: Same with this question, if someone could offer some advice on how to do this. I tried searching it but still was confusing.I'm still searching up youtube videos, but there aren't as helpful.

(comment also) #150: In the answer key it says -1 < x < 5. But in the problem it shows this: -1 < x < 5. It’s a small typo, just wanted to let you know Dr. Romano

#153: So each side length is sqrt(10). Could I have thought about this as being a 1:1:sqrt(2) triangle?? Bc at first that is what I thought. Which would be sqrt(10): sqrt(10) : 2 sqrt(10). ---But then I also did the Pythagorean theorem and got a different answer, only to see my initial ‘idea’ was wrong. Can you explain why we aren’t allowed to do that.


Sorry for the long post Dr. Romano and thank you for your help!

 

[goprodent2]

138: Yes probably a typo, should be y^3 = y^2 * y^1

146: I don't understand the first part either but this might help you for the second part:
(tanx + cotx)/(cscx) ---> ( (sinx/cosx) + (cosx/sinx) )/ (1/sinx) <-- do you get what I did ?
Then: combine the numerator (bolded) ----> ( (sin^2x + cos^2x)/ (cosxsinx) )/ (1/sinx)
The numerator of the fraction in the numerator (bolded) is an identity ---> becomes 1.
You now have: (1/cosxsinx)/(1/sinx)
Flip the bottom so that you get: (1/cosxsinx) * (sinx)
These cancel out (bolded) and you are left with 1/cosx which equals secx
It's hard to do it out like this lol if you need a picture let me know.

149: Logic --> so when you have an expansion, the sums of the exponents must equal the expansion exponent (ex. 2 + 3 = 5). You are multiplying two together, so their sum must be equal to 5 * 2 = 10. The only answer in which the sums of the exponents of the variables equals 10 is B.
There is the long Pascal's way of doing this, if you need that let me know.

153: You can either use the fact that a 45-45-90 triangle will have side lengths 1-1-sqrt2 or you can use the Pythagorean Theorem. Both will get you the same answer:
A=10, s= sqrt10 --> diagonal (d) = sqrt2 * s = sqrt2*sqrt10 = sqrt20 = 2sqrt5.
PT: s^2 + s^2 = d^2 ---> 10 + 10 = d^2 --> d^2 = 20 ---> d = sqrt20 ---> d = 2sqrt5

Hope this helps!

Thank you so much for the response. I made a note of #138.

I got 146. The wording was off and I couldn't follow the steps but everything that you did, I knew what to do and I get it know. Thank you!

for #149, you said "you are multiplying two together. so their sum must equal 5*2=10" I understand how you got 2+3 = 5 for the expansion exponent (one of exponent rules), What do you mean multiplying two together? Where did you get the 2? Does this "trick" always work? If possible could you show me how to do Pascal's way. If it's tough to show then no worries.

#153, I see my mistake when I did the 45-45-90. Slight error in my math. Awesome. glad both ways worked.

Thank you so much!! I really appreciate you taking the time to help!

 

[deleted768895]

Hello Dr Romano,

For Q.124 Organic Chemistry, DAT Destroyer 2017...Im struggling to understand what type of reaction this is? In the solutions it says to remove the alpha hydrogens, so why are the alpha hydrogens being protonated and why isn't the carbonyl carbon attacked?

Thanks!

 

[goprodent2]

DAT Destroyer OCHEM #172. When it asks which compound nitrates more rapidly? What does that mean? Does it mean which one can be substituted for a nairate?

Also, the answer is an activator, o/p director. Why are we looking for an activator? Couldn't we still use a meta director??

Thank you.

 

[DAT Destroyer]

Hello Dr Romano,

For Q.124 Organic Chemistry, DAT Destroyer 2017...Im struggling to understand what type of reaction this is? In the solutions it says to remove the alpha hydrogens, so why are the alpha hydrogens being protonated and why isn't the carbonyl carbon attacked?

Thanks!

This is called a Deuterium exchange reaction. In this reaction acidic protons (alpha Hydrogens) are being replaced by Deuterium ions. Small amount of DCL is used to protonate carbonyl oxygen, D2O is used to donate Deuterium ions.

Hope this helps.

 

[hsjordan24]

Dr. Romano,

For #116 in the gen chem destroyer 2017, I'm confused as why the Keq changes as a result of changing the concentration of reactants and products. I thought that only temperature could change K values in an equilibrium system? I understand that the Keq is the inverse for the reverse reaction. I guess I'm just missing something.

Thanks in advance.

 

[deleted768895]

This is called a Deuterium exchange reaction. In this reaction acidic protons (alpha Hydrogens) are being replaced by Deuterium ions. Small amount of DCL is used to protonate carbonyl oxygen, D2O is used to donate Deuterium ions.

Hope this helps.

Thank you!!

 

[deleted768895]

Dr. Romano,

For #116 in the gen chem destroyer 2017, I'm confused as why the Keq changes as a result of changing the concentration of reactants and products. I thought that only temperature could change K values in an equilibrium system? I understand that the Keq is the inverse for the reverse reaction. I guess I'm just missing something.

Thanks in advance.

Hey, I can attempt to answer this question for you...
So the point is that K is the equilibrium constant for a specific reaction. The reverse reaction is a new reaction and thus obtains a new constant (the reciprocal of the forward reaction). Because if you think about it, the reaction A + B -> AB would be expressed in eqm as [AB]/[A] and this is not the same reaction as AB -> A + B, which would be expressed as [A]/[AB].

Hope this helps clarify things!

 

[deleted768895]

Slight confused about the correct answer for Q16, DAT Destroyer QR, the correct answer is given at B which states that the area is doubled. But the solution says that the area is quadrupled?

 

[DAT Destroyer]

Dr. Romano,

For #116 in the gen chem destroyer 2017, I'm confused as why the Keq changes as a result of changing the concentration of reactants and products. I thought that only temperature could change K values in an equilibrium system? I understand that the Keq is the inverse for the reverse reaction. I guess I'm just missing something.

Thanks in advance.

For 116.....Only temperature will change the value of the Keq indeed. When you change the coefficients, the Keq APPEARS to change in value, but think of this as a " compensation " for the coefficient change. For example, in a DESTROYER question a reaction halves, thus Keq value appears to change by the square root. If a reaction doubles, the Keq appears to change by the square of its value. Essentially, we are only compensating for the coefficient change. In Physical Chemistry, this can be mathematically proven, but you don't need it. Bottom line......For your purposes.....only temperature will change the value of Keq. A coefficient change only appears to change the value of our equilibrium constant since the coefficients have changed but they are compensated for.

I hope this helps


Dr. Romano

 

[DAT Destroyer]

DAT Destroyer OCHEM #172. When it asks which compound nitrates more rapidly? What does that mean? Does it mean which one can be substituted for a nairate?

Also, the answer is an activator, o/p director. Why are we looking for an activator? Couldn't we still use a meta director??

Thank you.

For 172..... Activating Groups make a reaction go FASTER. Deactivating groups make a reaction go slower. Groups like OCH3 and CH3 represent activating groups. Groups like NO2 and COOH represent deactivating groups. The deactivating groups all deactivate the ring and direct meta. Activating groups will direct ortho and para. The Halogens and the nitroso..NO.... group will direct ortho-para but will slightly deactivate the ring. This questions essentially wants to know which group is the MOST activating. The OCH3 group is a more reactive group and more activating than a methyl or ethyl because in the mechanism there is more stabilization of the intermediate carbocation. Refer to any text to see this drawn out if you need further clarification. The Solomons text is best for this, but the Klein book can suffice.

I hope this helps.

Dr. Romano

 

[DAT Destroyer]

#186 OCHEM 2017:

I am not understanding how the monochlorination of 2-methylbutane would give 4 products (not counting stereoisomers). Why wouldn't the methyl group be able to be chlorinated? My guess is that by chlorinating this methyl group, it would yield the same exact product as chlorinating carbon 1, but not sure if that is correct

For 186.....Start by drawing out 2-methylbutane. OK.......number the carbons. Look at the methyl at carbon 1 and the methyl attached to carbon #2. These methyl groups are EQUIVALENT...thus when chlorinated, they give the same constitutional isomer. Draw it out and put a Cl on the methyl carbons and convince yourself they are equivalent by simply naming them.

I hope this helps.

Dr. Romano

 

[hsjordan24]

For 116.....Only temperature will change the value of the Keq indeed. When you change the coefficients, the Keq APPEARS to change in value, but think of this as a " compensation " for the coefficient change. For example, in a DESTROYER question a reaction halves, thus Keq value appears to change by the square root. If a reaction doubles, the Keq appears to change by the square of its value. Essentially, we are only compensating for the coefficient change. In Physical Chemistry, this can be mathematically proven, but you don't need it. Bottom line......For your purposes.....only temperature will change the value of Keq. A coefficient change only appears to change the value of our equilibrium constant since the coefficients have changed but they are compensated for.

I hope this helps


Dr. Romano

Yes, that makes sense thank you very much.

 

[joshthebear]

I am using the 2017 Destroyer to prepare for my DAT next week and I had a question on one of the Organic problems. On number 37 the solution says that choice B (Aniline) has an electron withdrawing group, therefore it can be eliminated, but from watching Chad's videos and my studies I was taught that an NH2 group on Benzene should be electron-donating, which should make Benzene more reactive, thus making it more basic. I was wondering if anyone could clear this up for me. Thank you!